a) ĐKXĐ: \(x\notin\left\{\frac{3}{2};0\right\}\)
Ta có: \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
\(\Leftrightarrow\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}-\frac{5\left(2x-3\right)}{x\left(2x-3\right)}=0\)
\(\Leftrightarrow x-3-5\left(2x-3\right)=0\)
\(\Leftrightarrow x-3-10x+15=0\)
\(\Leftrightarrow-9x+12=0\)
\(\Leftrightarrow-9x=-12\)
hay \(x=\frac{4}{3}\)(tm)
Vậy: \(x=\frac{4}{3}\)
c) ĐKXĐ: x∉{2;-2}
Ta có: \(\frac{x-1}{x-2}-\frac{x}{x-2}=\frac{5x-8}{x^2-4}\)
\(\Leftrightarrow\frac{1}{x-2}-\frac{5x-8}{x^2-4}=0\)
\(\Leftrightarrow\frac{x+2}{\left(x-2\right)\left(x+2\right)}-\frac{5x-8}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow x+2-\left(5x-8\right)=0\)
\(\Leftrightarrow x+2-5x+8=0\)
\(\Leftrightarrow10-4x=0\)
\(\Leftrightarrow4x=10\)
hay \(x=\frac{5}{2}\)(tm)
Vậy: \(x=\frac{5}{2}\)
